Lecture 21

Review

Recall the alternating series test from calculus: “Suppose $(a_n)^\infty_{n=1}$ is a sequence satisfies the following conditions:

1. The sequence is nonnegative. (For all $n\in \mathbb{N}$, $a_n\geq 0$.)
2. The sequence is decreasing. ($a_1\geq a_2\geq a_3\geq \cdots$)
3. $\lim_{n\to\infty}a_n=0$.

Then $\sum_{n=1}^\infty (-1)^{n+1}a_n$ converges.”

Exercise: Show that the statement above is false if we remove the second condition.

[Hint: Use the fact that $\sum_{n=1}^\infty \frac{1}{n}$ diverges.]

Let the sequence $a_n$ be defined as $a_n=\frac{1}{n},a_{n+1}=0$ for all $n\in \mathbb{N}$. This sequence satisfies the 1,3 but not the 2.

And the harmonic series is not convergent.

New Material

Other tests for convergence of series

Recall the integration by parts formula:

Let $A(t),a(t),b(t)$ be functions of $t$ and $A'(t)=a(t)$.

Then

Theorem 3.41 Summation by parts

Let $a_n,b_n$ be sequences.

Let $A(n)=\sum_{k=1}^n a_k$. ($A_{-1}=0$). If $0\leq p\leq q$, then

Proof:

QED

Theorem 3.42 (Dirichlet’s test)

Suppose

(a) the partial sum $A_n$ of $\sum a_n$ form a bounded sequence.
(b) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)
(c) $\lim_{n\to\infty}b_n=0$.

Then $\sum a_nb_n$ converges.

Proof:

By Cauchy criterion, it’s enough to prove

$\forall \epsilon >0, \exists N\in \mathbb{N}$ such that for all $p\geq q\geq N$,

By the partial sum $A_n$ of $\sum a_n$ form a bounded sequence. Let $\left|A_n\right|\leq M$ for all $n\in \mathbb{N}$.

Then we let $\epsilon >0$ be given. Since $b_n\to 0$, there exists $N\in \mathbb{N}$ such that for all $n\geq N$, $|b_n|<\frac{\epsilon}{2M}$.

If $q\geq p\geq N$, then

So $\sum a_nb_n$ converges.

QED

Theorem 3.43 (Alternating series test)

Let $(b_n)^\infty_{n=1}$ be a sequence such that:

(a) $b_1\geq b_2\geq b_3\geq \cdots$ (non-increasing) (b) $\lim_{n\to\infty}b_n=0$

Then $\sum_{n=1}^\infty (-1)^{n+1}b_n$ converges.

Proof:

Let $a_n=(-1)^{n+1}$

$A_n=\sum_{k=1}^n a_k=1$ if $n$ is odd, $0$ if $n$ is even.

So $|A_n|\leq 1$ for all $n\in \mathbb{N}$.

By Theorem 3.42, $\sum_{n=1}^\infty a_n b_n$ converges.

QED

Example:

Consider the power series $\sum_{n=0}^\infty \frac{z^n}{n}$.

The radius of convergence is $1$.

We claim that the series converges for all $z\in \mathbb{C}$ with $|z|=1$ and $z\neq 1$.

Theorem 3.44 Abel’s test

Let $(b_n)^\infty_{n=0}$ be a sequence such that:

(a) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing) (b) $\lim_{n\to\infty}b_n=0$

Then if $|z|=1$ and $z\neq 1$, $\sum_{n=0}^\infty b_nz^n$ converges.

Proof:

Fix $z\in \mathbb{C}$ with $|z|=1$ and $z\neq 1$.

Let $a_n=z^n$.

Then $A_n=\sum_{k=0}^n z^k=\frac{1-z^{n+1}}{1-z}$._

$|A_n|\leq \frac{|1-z^{n+1}|}{|1-z|}$ for all $n\in \mathbb{N}$.

By triangle inequality, $|1-z^{n+1}|\leq |1|+|z^{n+1}|=1+|z^{n+1}|$.

And since $|z|=1$, $|z^{n+1}|=|z|^{n+1}=1$.

So $|1-z^{n+1}|\leq 2$.

So $|A_n|\leq \frac{2}{|1-z|}$ for all $n\in \mathbb{N}$.

By Dirichlet’s test, $\sum_{n=0}^\infty b_nz^n$

QED

Absolute convergence

The series $\sum_{n=0}^\infty a_n$ is said to converge absolutely if $\sum_{n=0}^\infty |a_n|$ converges.

If $\sum_{n=0}^\infty a_n$ converges but does not converge absolutely, then $\sum_{n=0}^\infty a_n$ is said to converge conditionally.

Absolute convergence are nice but conditionally convergent series are not.

Theorem 3.45 (Absolute convergence)

If $\sum_{n=0}^\infty a_n$ converges absolutely, then $\sum_{n=0}^\infty a_n$ converges.

Proof:

Use comparison test.

QED

Rearrangement of series:

Let $f:\mathbb{N}\to \mathbb{N}$ be a bijection.

If $\sum_{n=0}^\infty a_n$ is a sequence and $b_n=a_{f(n)}$, then $(b_n)^\infty_{n=0}$ is a rearrangement of $(a_n)^\infty_{n=0}$.

If $\sum_{n=0}^\infty a_n$ converges absolutely, then any rearrangement of $\sum_{n=0}^\infty a_n$ converges to the same sum.

Example:

$a_n=\frac{(-1)^{n+1}}{n}$. $b_n=a_{f(n)}$.

You cannot always rearrange series.

But, if $\sum_{n=0}^\infty a_n$ converges absolutely, then you can rearrange the series.

Theorem 3.55

Let $(a_n)^\infty_{n=0}$ be a sequence in $\mathbb{C}$ such that $\sum_{n=0}^\infty |a_n|$ converges absolutely.

Then any rearrangement of $\sum_{n=0}^\infty a_n$ converges absolutely to the same sum.

Ideas of proof:

Let $f:\mathbb{N}\to \mathbb{N}$ be a bijection.

and let $b_n=a_{f(n)}$.

Let $s_n=\sum_{k=0}^n a_k,t_n=\sum_{k=0}^n b_k=\sum_{k=0}^n a_{f(k)}$.

$I_n=\{1,2,\cdots,n\}$.

$J_n=\{f(1),f(2),\cdots,f(n)\}$.

Key observation:

For every $n\in \mathbb{N}$, there exists a $p$ such that $\{1,2,\cdots,n\}\subset I_n\cap J_n$.

Then $|s_n-t_n|\leq \sum_{k=N+1}^\infty |a_k|$.

QED